COMPUTATIONAL PROBLEMS IN ABSTRACT ALGEBRA.

150 Marshall Hall Jr.
divides g, then also on qj (p- 1). These tests involving table look-ups were
easy to put on the machine.
The remaining 1146 orders were examined individually. The Stanton
principle, that if every set of characters for B&) (p dividing g to the first
power) contains a character of degree less than 2p, we must have w = 1,
could have been mechanized. But this test, illustrated in Example 3 of
0 5, was not difficult to apply by hand.
Certain orders were eliminated by consideration of Sylow groups of
order p2 or higher. Example 4 of 5 5 illustrates a relatively easy case.
The interplay between a Sylow p-group and a Sylow q-group provided
information in many cases. Since a group of order 35 is necessarily cyclic
it follows that if an S(5) is of order 5 and an S(7) is of order 7, then if 7
divides the order of N(5) then also 5 divides the order of N(7) and conversely.
Since an overwhelming fraction of the orders were divisible by a prime
p to exactly the first power, most investigations relied on the Brauer
theory (6) of modular characters for these. For the most part the computations
relied on the principal block B&). Here C(p) = S(p)X V(p). If
V(p) + 1 the restriction formulae (3.3), (3.4) and (3.5) were very valuable.
Example 5 eliminating the order g = 25200 = 16.9 -25 -7 illustrates several
of these principles, including the use of the Brauer-Fowler formula.
Example 6 shows how this theory is useful in constructing the simple
groups when they exist. For g = 29120 = 64.5.7.13, there is a unique
simple group, the Suzuki group Su(8). The Brauer theory makes the construction
of the complete character table easy. From this table we are
then able to deduce that G has a doubly transitive permutation representation
on 65 letters. We can then construct this permutation representation
and thus prove the existence and uniqueness of a simple group of this
order. Z. Janko [25] has shown how to use the character table of his group
to construct it as a matrix group of dimension 7 over GF(I1).
5. Examples of application of the general theory.
EXAMPLE 1. g is a multiple of p = 59. Here g = 59(1+ 59r)qw, q* 2.
Thus i < 143. In the formula (3.11) h < 4. r = F(59, u, h). There are 9 combinations
(h, u, r) satisfying (3.11).
(4 u, 4 (P-l)U+rp) = (l+upWp-11, g = (l+rplpqw
(1, 1, 31) 58.1830 = 60.1769, g = 107,970qw
(1, 28, 85) 58.5016 = 1653.176, g = 295,944qw
(1,57, 115) 58.6786 = 3364.117, g = 400,374qw
(2, 1, 61) 58.3600 = 60.3480, g = 212,400qw
(2, 3,w 58 * 5429 = 178.1769, g = 320,311qw
(2,28, 142) 58.8379 = 1653.294, g = 494,361qw
Simple groups of order less than one million 151
(3, 1,911 58.5370 = 60.5191, g = 316,830qw
(3,2, 121) 58.7140 = 119.3480, g = 421,260qw
(4, 1, 121) 58.7140 = 60.6902, g = 421,260qw
The last two give the same value of r so that we are dealing with a triple
factorization
58.7140 = 60.6902 = 119.3480.
In every case then we have qw < 9 and so necessarily q = 2. We have shown
that we need consider only values of g which are multiples of 168 or 48.
We consider the factorization of 1 frp.
1830 = 2.3.5.61
5016 = 8.3.11.19
6786 = 2.81.43
3600 = 16.9.25
5429 = 61.89
8379 = 9.49.19
5370 = 2.3.5.179
7140 = 4-3.5.7.17
In order that g be a multiple of 168 or 48, with the possible values of qw
making g -= l,OOO,OOO, we need consider only 1 +rp = 1830, q = 2, w = 4;
l+rp=5016, q=2, w = l ; l+rp=3600, q=z, w = l o r 2 ;
lfrp = 7140, q = 2w = 1.
The basic relations (3.2) on the degrees of &(59) reduce here to
1+84fo+Blfi = 0, S& El, Sofa G -2(59),
f0IW+rp), fij2U+rp), fo,fi=-29. (5.1)
For the four possible values for 1 frp we find degrees satisfying (5.1)
only in the first and last cases.
1830 S(59)‘s l-61+60 = 0 (5.2)
7140 S(59)‘s l-120f119 = 0
For 1830 S(59)‘s, since the degree 60 is less than 2p = 118, the Stanton
condition (8) requires w = 1, but this is in conflict with the condition
that g be a multiple of 4. Hence we may exclude 1830 S(59)‘s. The only
case remaining is that of 7140 S(59)‘s, q = 2, w = 1 and we have
g = 7140.59.2 = 842,520 = 8*3*5.7.17.59,
IiV(59)l = 118 = 2.59, q = 2, w = 1. (5.3)
Degrees in Bo(59): l-120f119 = 0.
Here a group of order 17.59 is necessarily Abelian and so since 177 IN(59)1
it follows that 59{jN(17) I. The only possible numbers of S(17)‘s are
CPA 11